Question

\({\bf{Work~and~Power}}\)

Master Page: https://questioncove.com/study#/updates/5b0354be2f45c9d5a6dd3800

Answer 1

Work is the product of a force exerted over some distance. Both of these quantities must be parallel to each other. For example:
\[W = F_{||} \times d\]
Where w = work, F|| = parallel force, and d = displacement

In order for work to occur, force and displacement must be parallel

Pushing a box across a table is work done in the horizontal direction, or along an x axis.

Lifting a cartoon of milk upwards is work done in the vertical direction, or along the y axis.

You cannot calculate the work done in the x axis with a force from the y axis. Example: a hockey puck's weight used as force whilst it is sliding across ice. Instead you would use friction and the initial push to determine the work done.

Quick notes:
-Work is a scalar property
-It can be both positive and negative
-Negative work occurs when force and displacement are in opposite directions.
-Positive work is when force and displacement are in the same direction
-The unit for work is the Joule

Power is the rate at which work is done
\[P = \frac{ W }{ t }\]

The units for power is a watt, w = J/s
1 horsepower = 746 watts (745.7)

Note the other forms of the formula for power:
\[P = \frac{ F \times d }{ t }\]
\[P = F \times V_{avg}\]
Remember the force and average velocity must be in the same direction.

Answer 2

Note that:
\[F_{g} = F_{gravitational} = mg = mass \times gravity\]
There is some altering between using different values for gravity (most notably 9.8 and 10).

35. The total work done in lifting a typical high school physics textbook a vertical distance of 0.10 meter is approximately
\[W = F_{||} \times d \]
\[W = mg \times h\]
\[W = 1.5(10)(0.10) = 1.5 J\]
Where the average mass of a high school physics textbook is 1.5 kg

36. Through what vertical distance is a 50-newton object moved if 250 joules of work is done against the gravitational field of Earth.
\[W = F_{||} \times d\]
\[250 = 50 \times d\]
\[d = 5m\]
37. A small electric motor is used to lift a 0.50-kilogram mass at a constant speed. if the amass is lifted a vertical distance of 1.5 meters in 5.0 seconds, the average power developed by the motor is
\[P = F \times V_{avg}\]
\[P = mg \times V_{avg}\]
\[P = 0.50(10)(\frac{ 1.5 }{ 5 }) = 1.5 W\]
38. What is the maximum amount of work that a 6000-watt motor can do in 10 seconds?
\[P = \frac{ W }{ t }\]
\[W = P \times t\]
\[W = 6000 \times 10 = 60000 = 6.0 \times 10^{4} J\]
39. How much work is done by the force lifting a 0.1-kilogram hamburger vertically upward at constant velocity 0.3 meter from a table?
\[W = F_{||} \times d\]
\[W = mg \times d\]
\[W = 0.1(10) \times 0.3 = 0.3 J\]
40. The watt times second is a unit of

The answer is energy. Look at problem #38 where power (watts) was multiplied by time (seconds). Thus, watts per second. Joules is energy.

41. Which quantity has both a magnitude and direction?

Impulse, which is force applied over some time. Force is a vector quantity, thus impulse is as well.

42. Two elevators, A and B, move at constant speed. Elevator B moves with twice the speed of elevator A. Elevator B weighs twice as much as elevator A. Compared to the power needed to lift elevator A, the power needed to lift elevator B is

It is the same because:
\[P = F_{||} \times V_{avg}\]
P = F_{||} times 2V_{avg}
P = 2F_{||} times V_{avg}
Both increase by a factor of two.

43. What is the maximum height to which a motor having a power rating of 20.4 watts can lift a 5.00-kilogram stone vertically in 10 seconds?
\[P = \frac{ F_{||} \times d }{ t }\]
\[d = \frac{ P \times t }{ F_{||}}\]
\[d = \frac{ 20.4 \times 10 }{5.00(9.8)} = 4.16 m\]